how to find indefinite integral of f(x)=sin^2(8x)

Question

how to find definite integral  of the following.

 

f(x)=  sin^2(8x)

Answer 1

cos2A=1-2sin^2A so sin^2A=(1-cos2A)/2. Put A=8x then sin^2(8x)=(1-cos(16x))/2.

Integral(sin^2(8x)dx)=integral((1/2)-(1/2)cos(16x)dx)=x/2-(1/32)sin(16x)+k where k is constant of integration, because no limits for a definite integral have been supplied.

Answer 2

Integral (sin^2(8x))dx

8x = z

8dx = dz

dx = dz/8

Integral(sin^2(8x))=Integral(sin^2z)dz/8=(1/8)Integral(sin^2z)dz

u = sinz               Integralsinzdz = dv

du = cosz          -cosz =v

Integral(sin^2z)dz = =( -sinz*cosz +Integral(cos^2z)dz))

cos^2z= 1-sin^2z

Integral(sin^2z)dz  = ( -sinz*cosz + Integral(1-sin^2z)dz)

Integral(sin^2z)dz =( -sinz*cosz +Integraldz -Integral(sin^2z)dz)

Integral(sin^2z)dz                                          Integral(sin^2z)dz

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2*Integral(sin^z)dz =( z -sinzcosz)

Integral(sin^2z)dz = z/2 -  sinz*cosz/2

Integral(sin^2z)dz = z/2 - (sin2z)/4

Integral(sin8x)dx = (1/8)(z/2- sin2z/4)= z/16 -( sin2z)/32

Integral(sin8x)dx = 8x/16 - (sin16x)/32 = x/2 - (sin16x)/32 + K